A) \[MgL\]
B) \[\frac{1}{3}MgL\]
C) \[\frac{1}{9}MgL\]
D) \[\frac{1}{18}MgL\]
Correct Answer: D
Solution :
If m is the mass per unit length of the chain, the mass of length y will be \[ym\]and the force acting on it due to gravity will be \[mgy\](assuming that y is the length of the chain hanging over the edge). So, the workdone in pulling the dy length of the chain on the table \[dW=F(-dy)\] (as\[y\]is decreasing) i.e., \[dW=(mgy)(-dy)\] (as \[F=mgy\]) So, the work done in pulling the hanging portion on the table \[W=-\int_{L/g}^{0}{mgy\,dy=\frac{mg{{L}^{2}}}{2{{(3)}^{2}}}}\] \[=\frac{mgL}{18}\] \[(as\,M=mL)\]You need to login to perform this action.
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